PHP variable scope - How to share variable between two functions without using arguments?

Question

I have a small problem in a script I can't solve on my own. Having studied the PHP documentation regarding variable scope, I am not sure if this is even possible.

Take the following example:

function my_funct_a() {
    // Do stuff
    return $a;
}

function my_funct_b() {
    // Do other stuff
    return $a + $b;
}

You see, the problem is that in my_funct_b, $a is not available because it was declared (and is returned) in my_funct_a.

Normally I would pass this variable as an argument, but this is not possible at this point due to some kind of framework limitation.

So I tried to do it like this:

function my_funct_a() {
    // Do stuff
    global $a;
    return $a;
}

function my_funct_b() {
    // Do other stuff
    global $a;
    return $a + $b;
}

This also didn't work, I think because global works 'the other way around'. Instead of declaring a variable as global inside a function to be available outside the function, it has to be declared as global outside the function to be available inside the function.

The Problem is that the value of $a is created in my_funct_a, so I can't global it before the value is known.
Because of that, I tried to do it like this:

// global variable, but no value assigned yet
global $a

function my_funct_a() {
   // Do stuff
   global $a;
   $a = 1; 
   return $a;
}

function my_funct_b() {
   // Do other stuff
   global $a;
   return $a + $b;
}

This also didn't work. Why? Is it even possible without passing the variable as an argument?

Answer 1

The best way to encapsulate data and functionality is a class:

class Foo {

    protected $val = 1;

    public function inc() {
        return $this->val += 1;
    }

    public function dec() {
        return $this->val -= 1;
    }

}

I'm showing this because it should be used in favour of global variables attempt.

However, this should work:

$val = 1;

function inc() {
    global $val;
    return $val += 1;
}


function dec() {
    global $val;
    return $val -= 1;
}

Answer 2

If you need the value of $a and it can only be created with my_funct_a(), then add this as a dependency inside my_funct_b():

function my_funct_b() 
{
    $a = my_funct_a();
    // Do other stuff

    return $a + $b;
}

Answer 3

You put the global declaration inside the function. That causes uses of the variable within that function to refer to a global variable, not a local one. And if two functions both access the same global variable, they'll see the same values.

function my_funct_a() {
    global $a;
    $a = 10;
    return $a;
}

function my_funct_b() {
    $b = 5;
    global $a;
    return $a + $b;
}

echo my_funct_a() . "\n";
echo my_funct_b();

DEMO

Answer 4

$a = 1;
$b = 2;

function Sum()
{
    global $a, $b;

    $b = $a + $b;
} 

Sum();
echo $b;

The above script will output 3. By declaring $a and $b global within the function, all references to either variable will refer to the global version.

http://www.php.net/manual/en/language.variables.scope.php